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3 years ago

15

please help on lesson 13.2 independent practice. Pierre's parents ordered some pizzas for a party. 4.5 pizzas were eaten at the

party. There were at least 5 1/2 whole pizzas left over. How many pizzas did Pierre's parents order? QUICKLY!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

2 answers:

Mariana [72]3 years ago

7 0

1/2 means .5 so 4.5+ 5.5= 10 pizzas were ordered.

Zigmanuir [339]3 years ago

4 0

10 because 4 1/2+ 5 1/2= 10

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3 years ago

Plz I need Answer For It Will give Good Rating

avanturin [10]

Answer:

B

Step-by-step explanation:

1. Find where the line crosses on the y-axis

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3 years ago

On Monday, during a one day sale, the price of a suit was decreased by 20%. On Tuesday, the price of the suit was changed back t

nignag [31]

Answer:

25%

Step-by-step explanation:

If we consider the original price of the piece of suit is x.

Given that, on Monday during a one day sale, the price of the suit was decreased by 20%.

Therefore, the new price of the suit on Monday is $x( 1 - \frac{20}{100}) = 0.8x$ .

Now, the price of the suit on Tuesday was changed back to the original price i.e. x.

Therefore, the price is increase by (x - 0.8x) = 0.2x from the price of 0.8x.

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3 years ago

Please help me for the love of God if i fail I have to repeat the class

Elena-2011 [213]

$\theta$ is in quadrant I, so $\cos\theta>0$ .

$x$ is in quadrant II, so $\sin x>0$ .

Recall that for any angle $\alpha$ ,

$\sin^2\alpha+\cos^2\alpha=1$

Then with the conditions determined above, we get

$\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35$

and

$\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}$

Now recall the compound angle formulas:

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$

$\sin2\alpha=2\sin\alpha\cos\alpha$

$\cos2\alpha=\cos^2\alpha-\sin^2\alpha$

as well as the definition of tangent:

$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}$

Then

1. $\sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}$

2. $\cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}$

3. $\tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}$

4. $\sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}$

5. $\cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}$

6. $\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=-\dfrac{24}7$

7. A bit more work required here. Recall the half-angle identities:

$\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2$

$\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2$

$\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}$

Because $x$ is in quadrant II, we know that $\dfrac x2$ is in quadrant I. Specifically, we know $\dfrac\pi2$ , so $\dfrac\pi4$ . In this quadrant, we have $\tan\dfrac x2>0$ , so

$\tan\dfrac x2=\sqrt{\dfrac{1-\cos x}{1+\cos x}}=\dfrac32$

8. $\sin3\theta=\sin(\theta+2\theta)=\dfrac{44}{125}$

6 0

4 years ago